3.707 \(\int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx\)

Optimal. Leaf size=171 \[ -\frac {(b c-a d)^2 (5 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{3/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d) (5 a d+b c)}{8 b^3 d}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{12 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d} \]

[Out]

-1/8*(-a*d+b*c)^2*(5*a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(7/2)/d^(3/2)-1/12*(5*a*d
+b*c)*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b^2/d+1/3*(d*x+c)^(5/2)*(b*x+a)^(1/2)/b/d-1/8*(-a*d+b*c)*(5*a*d+b*c)*(b*x+a)
^(1/2)*(d*x+c)^(1/2)/b^3/d

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Rubi [A]  time = 0.09, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {80, 50, 63, 217, 206} \[ -\frac {(b c-a d)^2 (5 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{3/2}}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{12 b^2 d}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d) (5 a d+b c)}{8 b^3 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[(x*(c + d*x)^(3/2))/Sqrt[a + b*x],x]

[Out]

-((b*c - a*d)*(b*c + 5*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^3*d) - ((b*c + 5*a*d)*Sqrt[a + b*x]*(c + d*x)^(3
/2))/(12*b^2*d) + (Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*b*d) - ((b*c - a*d)^2*(b*c + 5*a*d)*ArcTanh[(Sqrt[d]*Sqrt
[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(7/2)*d^(3/2))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx &=\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(b c+5 a d) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx}{6 b d}\\ &=-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {((b c-a d) (b c+5 a d)) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{8 b^2 d}\\ &=-\frac {(b c-a d) (b c+5 a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^3 d}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {\left ((b c-a d)^2 (b c+5 a d)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 b^3 d}\\ &=-\frac {(b c-a d) (b c+5 a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^3 d}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {\left ((b c-a d)^2 (b c+5 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{8 b^4 d}\\ &=-\frac {(b c-a d) (b c+5 a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^3 d}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {\left ((b c-a d)^2 (b c+5 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{8 b^4 d}\\ &=-\frac {(b c-a d) (b c+5 a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^3 d}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(b c-a d)^2 (b c+5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 153, normalized size = 0.89 \[ \frac {\sqrt {c+d x} \left (\sqrt {d} \sqrt {a+b x} \left (15 a^2 d^2-2 a b d (11 c+5 d x)+b^2 \left (3 c^2+14 c d x+8 d^2 x^2\right )\right )-\frac {3 (b c-a d)^{3/2} (5 a d+b c) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{24 b^3 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(c + d*x)^(3/2))/Sqrt[a + b*x],x]

[Out]

(Sqrt[c + d*x]*(Sqrt[d]*Sqrt[a + b*x]*(15*a^2*d^2 - 2*a*b*d*(11*c + 5*d*x) + b^2*(3*c^2 + 14*c*d*x + 8*d^2*x^2
)) - (3*(b*c - a*d)^(3/2)*(b*c + 5*a*d)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/Sqrt[(b*(c + d*x))/(
b*c - a*d)]))/(24*b^3*d^(3/2))

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fricas [A]  time = 0.88, size = 412, normalized size = 2.41 \[ \left [\frac {3 \, {\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{2} + 3 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{4} d^{2}}, \frac {3 \, {\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} + 3 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{4} d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(b^3*c^3 + 3*a*b^2*c^2*d - 9*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c
*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(8
*b^3*d^3*x^2 + 3*b^3*c^2*d - 22*a*b^2*c*d^2 + 15*a^2*b*d^3 + 2*(7*b^3*c*d^2 - 5*a*b^2*d^3)*x)*sqrt(b*x + a)*sq
rt(d*x + c))/(b^4*d^2), 1/48*(3*(b^3*c^3 + 3*a*b^2*c^2*d - 9*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(-b*d)*arctan(1/2*(2
*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) +
2*(8*b^3*d^3*x^2 + 3*b^3*c^2*d - 22*a*b^2*c*d^2 + 15*a^2*b*d^3 + 2*(7*b^3*c*d^2 - 5*a*b^2*d^3)*x)*sqrt(b*x + a
)*sqrt(d*x + c))/(b^4*d^2)]

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giac [B]  time = 1.25, size = 346, normalized size = 2.02 \[ \frac {\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} + \frac {b^{6} c d^{3} - 13 \, a b^{5} d^{4}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{7} c^{2} d^{2} + 2 \, a b^{6} c d^{3} - 11 \, a^{2} b^{5} d^{4}\right )}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b d^{2}}\right )} d {\left | b \right |}}{b^{2}} + \frac {6 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )} c {\left | b \right |}}{b^{3}}}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/24*((sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*c*d^3 - 13*a*b^5
*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2*a*b^6*c*d^3 - 11*a^2*b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*
a^2*b*c*d^2 - 5*a^3*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b
*d^2))*d*abs(b)/b^2 + 6*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x +
a) + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b
*d)))/(sqrt(b*d)*d))*c*abs(b)/b^3)/b

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maple [B]  time = 0.02, size = 395, normalized size = 2.31 \[ -\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (15 a^{3} d^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-27 a^{2} b c \,d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+9 a \,b^{2} c^{2} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 b^{3} c^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-16 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} d^{2} x^{2}+20 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x -28 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c d x -30 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2}+44 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d -6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(d*x+c)^(3/2)/(b*x+a)^(1/2),x)

[Out]

-1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(-16*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*d^2*x^2+15*a^3*d^3*ln(1/2*(2*b*
d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))-27*a^2*b*c*d^2*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x
+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))+9*a*b^2*c^2*d*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(
b*d)^(1/2))/(b*d)^(1/2))+3*b^3*c^3*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))
+20*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*d^2*x-28*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c*d*x-30*(b*d)^(1
/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2+44*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d-6*(b*d)^(1/2)*((b*x+a)*(d*x
+c))^(1/2)*b^2*c^2)/b^3/d/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\left (c+d\,x\right )}^{3/2}}{\sqrt {a+b\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(c + d*x)^(3/2))/(a + b*x)^(1/2),x)

[Out]

int((x*(c + d*x)^(3/2))/(a + b*x)^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d*x+c)**(3/2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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